Hey All!

Please let me know I am included within the accepted organization or not?

According to Karachi, Pakistan timezone last night was my GSoC'23 result.

Please guide me through the procedure how to check my GSoC'23 result.

Name :- Sumaiya Qureshi (Karachi, Pakistan)

I opted for Improvements to mathematics interaction with the desired
organization "Oppia" within Web category.

Let me know if you all sage developers want further information regarding
myself to search my result.

Regards,
Sumaiya.



On Wed, May 3, 2023, 5:19 PM Georgi Guninski <[email protected]> wrote:

> Cryptography based on hardness of finding solutions of diophantine
> equations.
>
> This is related to crypto and there is money in crypto,
> so someone may profit :)
>
> With latex on mathoverflow [1].
> For the general approach, check [2]
>
> Working over $K=\mathbb{Q}[x_1,...,x_n,y_1,...y_m]$.
>
> Let $f_i=g_i \cdot (h_i(y_i)+l_i(x_i))$ where $l_i$ is linear
> and depends on only $x$ variables and $h_i$ depends on only $y$
> variables.
> There are no restrictions on $g_i$.
>
> Let $F=\sum_{i=1}^k f_i$ be given as sum of monomials.
>
> If we know the set of $f_i$ (the secret trapdoor) and we are given $y_i$
> , for sufficiently general $l_i$, we can find the solutions
> of $F=0$ as the solutions of the linear in $x_i$ system
> of linear equations $(h_i(y_i)+l_i(x_i))=0$
>
> >Q1 Can we find $f_i$ such that solving $F=0$ is hard
> without knowing the trapdoor?
>
> >Q2 Assume an oracle gives many solutions to $F=0$,
> can we still get hardness results for new solutions?
>
>
> [1] Complexity of finding solutions of trapdoored polynomial
>
> https://mathoverflow.net/questions/445899/complexity-of-finding-solutions-of-trapdoored-polynomial
>
> [2] Cryptography signature scheme based on hardness of finding points
> on varieties?
>
> https://mathoverflow.net/questions/445898/cryptography-signature-scheme-based-on-hardness-of-finding-points-on-varieties
>
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